What is the extraneous solution to these equations? $\dfrac{x^2 - x}{x + 9} = \dfrac{-7x + 27}{x + 9}$
Solution: Multiply both sides by $x + 9$ $ \dfrac{x^2 - x}{x + 9} (x + 9) = \dfrac{-7x + 27}{x + 9} (x + 9)$ $ x^2 - x = -7x + 27$ Subtract $-7x + 27$ from both sides: $ x^2 - x - (-7x + 27) = -7x + 27 - (-7x + 27)$ $ x^2 - x + 7x - 27 = 0$ $ x^2 + 6x - 27 = 0$ Factor the expression: $ (x + 9)(x - 3) = 0$ Therefore $x = -9$ or $x = 3$ At $x = -9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -9$, it is an extraneous solution.